Problem: Simplify the following expression and state the condition under which the simplification is valid. You can assume that $z \neq 0$. $y = \dfrac{6z}{9(3z + 1)} \div \dfrac{z}{4(3z + 1)} $
Dividing by an expression is the same as multiplying by its inverse. $y = \dfrac{6z}{9(3z + 1)} \times \dfrac{4(3z + 1)}{z} $ When multiplying fractions, we multiply the numerators and the denominators. $y = \dfrac{ 6z \times 4(3z + 1) } { 9(3z + 1) \times z } $ $ y = \dfrac{24z(3z + 1)}{9z(3z + 1)} $ We can cancel the $3z + 1$ so long as $3z + 1 \neq 0$ Therefore $z \neq -\dfrac{1}{3}$ $y = \dfrac{24z \cancel{(3z + 1})}{9z \cancel{(3z + 1)}} = \dfrac{24z}{9z} = \dfrac{8}{3} $